Q.Evaluate: $\displaystyle\lim_{x\to 1}\dfrac{x^{2}-1}{x^{3}-1}$

Direct substitution: $\lim_{x\to4}\sqrt{4x+9}=\sqrt{25}=5$.

Since $4x+9>0$ near $x=4$, the square-root function is continuous there and the limit is obtained by direct substitution:

$\lim_{x\to4}\sqrt{4x+9}=\sqrt{4(4)+9}=\sqrt{16+9}=\sqrt{25}=5.$

Substituting $x = -3$ into $y = 10 - 3x$ gives $y = 19$ — option (c).

GSEB Class-12 Statistics, Limit chapter:

Since $y = 10 - 3x$ is a polynomial (linear) function, it is continuous everywhere, so the limit is found by direct substitution:

$\lim_{x \to -3} (10 - 3x) = 10 - 3(-3) = 10 + 9 = 19$

Substituting $x = -1$: $-4 + k = 6$, so $k = 10$.

GSEB Class-12 Statistics, Limit chapter:

Since $4x + k$ is a linear (continuous) function, the limit equals its value at $x = -1$:

$\lim_{x \to -1}(4x + k) = 4(-1) + k = -4 + k$

Setting this equal to the given limit $6$:

$-4 + k = 6$

$k = 6 + 4 = 10$

Direct substitution gives $\lim_{x\to3}(3x-1)=3(3)-1=8$. Option (D).

Since $3x-1$ is a continuous polynomial function, the limit at $x=3$ is obtained by direct substitution:

$\lim_{x\to3}(3x-1)=3(3)-1=9-1=8.$

The limit of a constant is that constant, so $\lim_{x\to -2} 10 = \mathbf{10}$.

Reasoning. For any constant $c$, $\lim_{x\to a} c = c$, since the function value never changes as $x$ varies. Here $c=10$.

Using the standard limit $\lim_{x\to a}\dfrac{x^n-a^n}{x-a}=n\,a^{n-1}$ with $n=4,\ a=3$: value $=4\cdot 27 = \mathbf{108}$.

Given: $\lim_{x\to 3}\dfrac{x^4-81}{x-3}$. At $x=3$ it is the indeterminate form $\dfrac{0}{0}$.

Method (factorisation). Note $81 = 3^4$, so

$x^4-81 = (x^2-9)(x^2+9) = (x-3)(x+3)(x^2+9).$

Cancel $(x-3)$:

$\lim_{x\to 3}(x+3)(x^2+9) = (3+3)(9+9) = 6\times 18 = 108.$

(Equivalently $n\,a^{n-1}=4\cdot 3^3 = 108$.)

Direct substitution: $\lim_{x\to 3}(3x-1) = 3(3)-1 = 8$ — option (d).

A polynomial function is continuous for all real $x$, so its limit equals its value at the point:

$\lim_{x \to 3}(3x - 1) = 3(3) - 1 = 9 - 1 = 8$